epsilon delta of a limit

Question

Hey can anybody help me with the following proof? I am trying to solve the following limit using epsilon delta and I have found the limit to be 1/3 using the squeeze theorem and have got to this thus far but am a bit confused where I go now as I have both a 3x and a sinx when trying to find an epsilon?? Thanks in advance!!

Answer 1

You can easily show that $$\frac{\sin(x)+1}{3x+1}<\frac{2}{3x}$$ and then it is easy to show that $$\frac{2}{3x} < \epsilon $$

Answer 2

You are using the wrong definition of limits for this case. When dealing with limits at infinity, you want to use this definition. As RowanS stated, you can use the fact that the absolute value of sin(x) is bounded in the proof.